Proofs Regarding Primorial Patterns by Dennis R. Martin DP Technology Corp., Camarillo, CA [email protected] BSME, Michigan Technological University, Houghton, MI Department of Mathematics, University of California, Santa Barbara [email protected] Copyright © 2006 by Dennis R. Martin ALL RIGHTS RESERVED No part of this document may be reproduced, retransmitted, or redistributed by any means, without the expressed written consent of Dennis R. Martin. This document is available in other formats. See www.primenace.com. Abstract Observations are made regarding the pattern of the composite numbers that have a particular prime factor for their lowest prime factor. It is subsequently proven that this pattern repeats over intervals equal to the primorial of that lowest prime factor such that the number and distribution of such composites is constant. The value of that constant composite to primorial ratio is proven to be related to the previous prime numbers and its constant composite to primorial ratio. !1 1 Introduction It seems well known that the pattern of composite numbers repeats with a period equal to the primorial of each prime factor. Dickson [1] refers to remarks by H.J.S Smith in 1857 and papers by J. DeChamps published in 1907 regarding this property, and Weisstein [2] makes note of it. However, actual proofs of these properties do not seem to be readily available. This paper attempts to rectify that situation by starting from the ground up and deriving some general relationships and developing proofs based on them. 2 Singular Identities A prime number is a number greater than 1 that has no positive integer divisors other than 1 and itself.[3] By the fundamental theorem of arithmetic, every positive integer greater than 1 can be uniquely represented by its prime divisors in what is called a prime factorization.[4, 5] Since a prime pN has no other positive integer divisors besides 1 and itself, the prime factorization of pN [6] is simply pN. A positive integer greater than 1 which is not prime is a composite number. By convention the number 1 is considered neither prime nor composite. Every positive integer greater than 1 is either a prime number or a composite number. For every positive integer greater than 1, then, one of the following statements must be true: 1. The prime pN is a factor in its prime factorization. 2. The prime pN is not a factor in its prime factorization. i If a number n has a prime factorization where pN is a single factor, that is pN is a factor and i = 1, 1 and if pN is the only factor other than 1, then n = pN and it is prime. Otherwise it is a composite number which we can label C. Lemma 1: For every composite number C having pN as a factor, one of the following statements must be true: 1. The lowest prime factor of C is smaller than pN. That is, pN-J is the lowest prime factor and pN-J < pN. i 2. The lowest prime factor is pN and it is the lone prime factor, i.e. pN = C where i " 2. 3. pN is the lowest prime factor while the other factor is a larger prime pN+J > pN or a k product of one or more larger primes such as (pN+J) or (pN+J) • (pN+L) or combinations of their higher powers. i Proof: If pN is a factor by itself, then it is of the form pN , but i cannot be 1 because then the number would be prime, so i must be greater than 1 and thus satisfy condition 2. If pN is not a factor by itself, then it must be combined with at least one other prime. If any of the other primes !2 © 2006 Dennis R. Martin All Rights Reserved in the prime factorization is less than pN then condition 1 is satisfied, otherwise all of them must be greater than pN, in which is the case covered by condition 3. Q.E.D. !3 © 2006 Dennis R. Martin All Rights Reserved 3 Primorial Soup The primorial is analogous to a factorial applied to the sequence of prime numbers.[7] The primorial for the prime pN is the product of all primes up to and including pN, and it is denoted as pN#. By the definition of factorial, p1# = 2, and then for every prime greater than 2: pN# = pN • pN–1# (1) What happens, though, if instead of just multiplying pN–1# by pN to get the next primorial value, we multiply all of the positive integers in the interval up to and including pN–1# by pN? What can we say about these numbers, and what about the other integers that we might need to fill in the new interval up to and including pN#? What would happen if we were to add some multiple of pN# onto all of them? What can we say about the lowest prime factors of those numbers? For example, start with the number 1, and multiply it by the first prime p1 = 2. We now have an interval the width of the primorial p1# = 2 containing the numbers 1 and 2. Add any multiple of 2 onto these two numbers. That is, let n be a non-negative integer where n = 0, 1, 2, 3, 4, 5… The result of the addition is an interval of width p1# = 2 containing the numbers 2n + 1 and 2n + 2. For those numbers, the statements that follow, where ≡ indicates congruence, are always true.[8] (2n + 1) ≡ 1 (mod 2) (2n + 2) ≡ 0 (mod 2) Therefore within every interval of width p1# = 2 we have one number that has p1 = 2 for its lowest prime factor and another number that does not have 2 for a factor. Except for the case where n = 0 and that other number is 1, that other number must either be prime or it must be a composite that has a higher prime pN+J > p1 for its lowest prime factor. Essentially we have just found that every even number is evenly divisible by 2 and that every odd number is not. Now take that first primorial and multiply it by the second prime, p2 = 3 and fill in the spaces in between. The result is an interval of width p2# = 6 containing the numbers 1, 2, 3, 4, 5, 6. Add any multiple of 6 on to those numbers. The result will be an interval the width of the primorial p2# = 6, still, containing the following numbers: (6n + 1), (6n + 2), (6n + 3), (6n + 4), (6n + 5), (6n + 6) Let us factor these values. While we could factor a 3 from out of two of those numbers, though, let us instead only factor the lowest prime factor possible out of each. This factoring produces: (6n + 1) ≡ 1 (mod 6) 2 • (3n + 1) ≡ 2 (mod 6) ≡ 0 (mod 2) 3 • (2n + 1) ≡ 3 (mod 6) ≡ 0 (mod 3) !4 © 2006 Dennis R. Martin All Rights Reserved 2 • (3n + 2) ≡ 4 (mod 6) ≡ 0 (mod 2) (6n + 5) ≡ 5 (mod 6) 2 • (3n + 3) ≡ 0 (mod 6) ≡ 0 (mod 2) Performing a lowest prime factorization like this allows us to more easily count the numbers within the primorial interval in terms of their lowest prime factor. Out of every p2# primorial interval, we can see that there are three numbers that have 2 for their lowest prime factor, one number that has 3 for its lowest prime factor (the 3 • (2n + 1) term), and two other numbers which do not have 2 or 3 as a factor at all. That there are three numbers that have 2 for their lowest prime factor makes sense because we multiplied the previous primorial by 3 and that primorial interval had one number that had 2 for its lowest prime factor. But notice also that the term that does have 3 for its lowest prime factor has the same form as the term that did not have 2 as a factor in that previous p1# primorial interval. That is: (2n + 1) ≡ 1 (mod 2) → 3 • (2n + 1) ≡ 3 (mod 6) ≡ 0 (mod 3) To summarize what can be concluded so far: a. 1 out of every 2 and 3 out of every 6 numbers have 2 for their lowest prime factor. b. 1 out of every interval of 6 numbers has 3 for its lowest prime factor. c. 4 numbers total out of every 6 have either 2 or 3 for their lowest prime factor. d. 2 out of every 6 numbers do not have 2 or 3 as a factor. Ignoring the trivial case where one of those is 1, those two numbers individually are either prime or have a prime that is higher than 3 for their lowest prime factor. The count in (d) can be calculated as the primorial value minus the count from (a) that have 2 for their lowest prime factor minus the count from (b) that has 3 for its lowest prime factor. But the count from (a) corresponds back to the count in the p1# = 2 interval that had 2 for a lowest prime factor and the count from (b) matches the count from the previous p1# = 2 interval which did not have 2 as a factor. Thus the count in (d) is directly related to counts in the previous primorial. 4 A Preliminary Proof by Induction By induction then, if we multiply the first interval of 6 by p3 = 5 and then add any multiple of p3# = 5 • 6 = 30, we should expect the three numbers that had 2 as their lowest prime factor in the previous p2# = 6 primorial to lead to 3 • 5 = 15 = 30 / 2 which have 2 as their lowest prime factor in each p3# = 30 primorial interval.